`lim_(xto0^(-)) (sum_(r=1)^(2n+1)[x^(r)]+(n+1))/(1+[x]+|x|+2x),` where `ninN` and `[.]` denotes the greatest integer function, equals

Explanation

To solve the limit

\[ \lim_{x \to 0^-} \frac{\sum_{r=1}^{2n+1} [x^r] + (n+1)}{1 + [x] + |x| + 2x}, \]

where \( n \in \mathbb{N} \) and \([.]\) denotes the greatest integer function, we will follow these steps:

### Step 1: Analyze the numerator

The numerator consists of the sum of the greatest integer function applied to \( x^r \) for \( r \) from 1 to \( 2n + 1 \), plus \( n + 1 \).

As \( x \to 0^- \) (approaching 0 from the left), \( x \) is negative and very close to 0.

- For \( r = 1 \), \([x^1] = [x] = -1\) (since \( x \) is negative).
- For \( r = 2 \), \([x^2] = [x^2] = 0\) (since \( x^2 \) is positive and less than 1).
- For odd \( r \) (like \( r = 3, 5, \ldots, 2n + 1 \)), \([x^r] = -1\).
- For even \( r \) (like \( r = 2, 4, \ldots, 2n\)), \([x^r] = 0\).

Thus, we can summarize the contributions:
- The odd powers contribute \(-1\) for \( n+1 \) odd terms.
- The even powers contribute \(0\) for \( n \) even terms.

The total contribution from the odd powers (which are \( n + 1 \) terms) is:
\[ -n \quad \text{(since there are \( n \) odd terms)} \]

Adding \( n + 1 \) gives:
\[ -n + (n + 1) = 1. \]

### Step 2: Analyze the denominator

The denominator is:
\[ 1 + [x] + |x| + 2x. \]

As \( x \to 0^- \):
- \([x] = -1\),
- \(|x| = -x\),
- \(2x\) is also negative.

Thus, substituting these values:
\[ 1 - 1 - x + 2x = 1 + x. \]

### Step 3: Combine the results

Now substituting back into the limit:
\[ \lim_{x \to 0^-} \frac{1}{1 + x}. \]

As \( x \to 0^- \), \( 1 + x \to 1 \).

### Final Result

Thus, the limit evaluates to:
\[ \frac{1}{1} = 1. \]

So, the final answer is:
\[ \lim_{x \to 0^-} \frac{\sum_{r=1}^{2n+1} [x^r] + (n+1)}{1 + [x] + |x| + 2x} = 1. \]

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